Difference between revisions of "1963 AHSME Problems/Problem 13"
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Assume <math>c,d \ge 0</math>, and WLOG, assume <math>a<0</math> and <math>a \le b</math>. This also takes into accout when <math>b</math> is negative. That means | Assume <math>c,d \ge 0</math>, and WLOG, assume <math>a<0</math> and <math>a \le b</math>. This also takes into accout when <math>b</math> is negative. That means | ||
− | <cmath>\frac{1}{2^-a} + 2^b = 3^c + 3^d</cmath> | + | <cmath>\frac{1}{2^{-a}} + 2^b = 3^c + 3^d</cmath> |
− | Multiply both sides by <math>2^-a</math> to get | + | Multiply both sides by <math>2^{-a}</math> to get |
− | <cmath>1 + 2^{-a+b} = 2^-a (3^c + 3^d)</cmath> | + | <cmath>1 + 2^{-a+b} = 2^{-a} (3^c + 3^d)</cmath> |
− | Note that both sides are integers. If <math>a \ne b</math>, then the right side is even while the left side is odd, so equality can not happen. If <math>a = b</math>, then <math>2^-a (3^c + 3^d) = 2</math>, and since <math>a<0</math>, <math>a = -1</math> and <math>3^c + 3^d = 1</math>. No nonnegative value of <math>c</math> and <math>d</math> works, so equality can not happen. Thus, <math>a</math> and <math>b</math> can not be negative when <math>c,d \ge 0</math>. | + | Note that both sides are integers. If <math>a \ne b</math>, then the right side is even while the left side is odd, so equality can not happen. If <math>a = b</math>, then <math>2^{-a} (3^c + 3^d) = 2</math>, and since <math>a<0</math>, <math>a = -1</math> and <math>3^c + 3^d = 1</math>. No nonnegative value of <math>c</math> and <math>d</math> works, so equality can not happen. Thus, <math>a</math> and <math>b</math> can not be negative when <math>c,d \ge 0</math>. |
Assume <math>a,b \ge 0</math>, and WLOG, assume <math>c < 0</math> and <math>c \le d</math>. This also takes into account when <math>d</math> is negative. That means | Assume <math>a,b \ge 0</math>, and WLOG, assume <math>c < 0</math> and <math>c \le d</math>. This also takes into account when <math>d</math> is negative. That means | ||
− | <cmath>2^a + 2^b = \frac{1}{3^-c} + 3^d</cmath> | + | <cmath>2^a + 2^b = \frac{1}{3^{-c}} + 3^d</cmath> |
− | Multiply both sides by <math>3^-c</math> to get | + | Multiply both sides by <math>3^{-c}</math> to get |
− | <cmath>3^-c (2^a + 2^b) = 1 + 3^{d-c}</cmath> | + | <cmath>3^{-c} (2^a + 2^b) = 1 + 3^{d-c}</cmath> |
That makes both sides integers. The left side is congruent to <math>0</math> modulo <math>3</math> while the right side is congruent to <math>1</math> or <math>2</math> modulo <math>3</math>, so equality can not happen. Thus, <math>c</math> and <math>d</math> can not be negative when <math>a,b \ge 0</math>. | That makes both sides integers. The left side is congruent to <math>0</math> modulo <math>3</math> while the right side is congruent to <math>1</math> or <math>2</math> modulo <math>3</math>, so equality can not happen. Thus, <math>c</math> and <math>d</math> can not be negative when <math>a,b \ge 0</math>. | ||
Assume <math>a,c < 0</math>, and WLOG, let <math>a \le b</math> and <math>c \le d</math>. This also takes into account when <math>b</math> or <math>d</math> is negative. That means | Assume <math>a,c < 0</math>, and WLOG, let <math>a \le b</math> and <math>c \le d</math>. This also takes into account when <math>b</math> or <math>d</math> is negative. That means | ||
− | <cmath>\frac{1}{2^-a} + 2^b = \frac{1}{3^-c} + 3^d</cmath> | + | <cmath>\frac{1}{2^{-a}} + 2^b = \frac{1}{3^{-c}} + 3^d</cmath> |
− | Multiply both sides by <math>2^-a \cdot 3^-c</math> to get | + | Multiply both sides by <math>2^{-a} \cdot 3^{-c}</math> to get |
− | <cmath>3^-c (1 + 2^{b-a}) = 2^-a (1 + 3^{d-c})</cmath> | + | <cmath>3^{-c} (1 + 2^{b-a}) = 2^{-a} (1 + 3^{d-c})</cmath> |
That makes both sides integers. The left side is congruent to <math>0</math> modulo <math>3</math> while the right side is congruent to <math>1</math> or <math>2</math> modulo <math>3</math>, so equality can not happen. Thus, <math>a</math> and <math>c</math> can not be negative. | That makes both sides integers. The left side is congruent to <math>0</math> modulo <math>3</math> while the right side is congruent to <math>1</math> or <math>2</math> modulo <math>3</math>, so equality can not happen. Thus, <math>a</math> and <math>c</math> can not be negative. | ||
Revision as of 16:18, 15 July 2018
Problem
If , the number of integers which can possibly be negative, is, at most:
Solution
Assume , and WLOG, assume and . This also takes into accout when is negative. That means Multiply both sides by to get Note that both sides are integers. If , then the right side is even while the left side is odd, so equality can not happen. If , then , and since , and . No nonnegative value of and works, so equality can not happen. Thus, and can not be negative when .
Assume , and WLOG, assume and . This also takes into account when is negative. That means Multiply both sides by to get That makes both sides integers. The left side is congruent to modulo while the right side is congruent to or modulo , so equality can not happen. Thus, and can not be negative when .
Assume , and WLOG, let and . This also takes into account when or is negative. That means Multiply both sides by to get That makes both sides integers. The left side is congruent to modulo while the right side is congruent to or modulo , so equality can not happen. Thus, and can not be negative.
Putting all the information together, none of can be negative, so the answer is .
See Also
1963 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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All AHSME Problems and Solutions |
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